Question

witbout actually finding zeroes of polynomial f(x)=3(x^2-1)+2x-5 , find sum and product of zeroes

Answer 1

sum of zeros = - coefficient of x/coefficient of x^2= -2/3
product of zeros= constant term/ coefficient of x^2= -8/3
Hope this helps

Answer 2

Answer:

sum of the zeroes =-b/a

=-2/3

product of the zeroes =c/a

=-8/3

Step-by-step explanation:

f(x)=3(x^2-1)+2x-5

= 3x^2-3+2x-5

= 3x^2+2x-8

a=3 ;b=2 ;c=-8

sum of the zeroes =-b/a

=-2/3

product of the zeroes =c/a

=-8/3